I was studying a proof of the Central Limit Theorem (for iid variables), and at some step, the proof requires me to find the limit of $$\lim_{n \to \infty} \left( 1-\frac{t^2}{2n}+o(\frac{t^2}{n}) \right)^n=e^{-\frac{t^2}{2}} \qquad(1)$$ where $o(x^k)$ is a function such that $\lim_{x\to\infty} \frac {o(x^k)}{x^k}=0$ for any positive integer k (little-o notation).
I know that, by the definition of $e$, we have: $$\lim_{n \to \infty} \left( 1-\frac{t^2}{2n} \right)^n=e^{-\frac{t^2}{2}}$$
Also, we can rearrange the terms in (1) to have a similar expression: $$\lim_{n \to \infty} \left( 1-a\frac{t^2}{2n} \right)^n$$ With $a=1-2\frac{o(\frac{t^2}{n})}{\frac{t^2}{n}}$.
However, I don't know how to argue that this expression must converge to $e^{-\frac{t^2}{2}}$. In an intuitive level, it seems logical that this must be the limit, though I would want to have a more rigourous argument. How can I calculate this limit?
$\ln(1+u)= u + o(u)$ when $u\to0$. So : $$\ln\left(1-\frac{t^2}{2n}+o\left(\textstyle\frac{1}{n}\right)\right) = -\frac{t^2}{2n}+o\left(\textstyle\frac{1}{n}\right)$$ and $$\left(1-\frac{t^2}{2n}+o\left(\textstyle\frac{1}{n}\right)\right)^n = e^{n\ln\left(1-\frac{t^2}{2n}+o\left(\frac{1}{n}\right)\right)} = e^{-\frac{t^2}{2}+o(1)} \xrightarrow[n\to\infty]{} e^{-\frac{t^2}{2}}$$ $o(1)$ meaning some sequence with limit $0$.